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GED Math : GED Math

Study concepts, example questions & explanations for GED Math

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All GED Math Resources

4 Diagnostic Tests 263 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #35 : Solving For The Variable

Solve for the variable: $\displaystyle x-7 = 2x-3$

Possible Answers:

$\displaystyle -4$

$\displaystyle 6$

$-\frac{10}{3}$ $\displaystyle -\frac{10}{3}$

$\displaystyle -10$

$-\frac{4}{3}$ $\displaystyle -\frac{4}{3}$

Correct answer:

$\displaystyle -4$

Explanation:

Subtract $\displaystyle x$ from both sides.

$\displaystyle x-7 -x= 2x-3-x$

$\displaystyle -7 = x-3$

Add 3 on both sides.

$\displaystyle -7 +3= x-3+3$

x=-4 $\displaystyle x=-4$

The answer is: $\displaystyle -4$

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Example Question #32 : Solving For The Variable

Give the solution set:

$\displaystyle 9 (x + 18) < 261$

Possible Answers:

$\left \{ x| x< 27\right \}$ $\displaystyle \left \{ x| x< 27\right \}$

$\left \{ x| x< 11 \right \}$ $\displaystyle \left \{ x| x< 11 \right \}$

$\left \{ x| x< 31\right \}$ $\displaystyle \left \{ x| x< 31\right \}$

$\left \{ x| x< 47\right \}$ $\displaystyle \left \{ x| x< 47\right \}$

Correct answer:

$\left \{ x| x< 11 \right \}$ $\displaystyle \left \{ x| x< 11 \right \}$

Explanation:

First, distribute the 9 on the left by multiplying it by each expression in the parentheses:

$\displaystyle 9 (x + 18) < 261$

$9 \cdot x + 9 \cdot 18 < 261$ $\displaystyle 9 \cdot x + 9 \cdot 18 < 261$

$\displaystyle 9 x +162 < 261$

Isolate $\displaystyle x$ on the right by first, subtracting 162 from both sides:

$\displaystyle 9 x +162 - 162 < 261 - 162$

$\displaystyle 9 x < 99$

Divide both sides by 9:

$\frac{9x }{9}< \frac{99}{9}$ $\displaystyle \frac{9x }{9}< \frac{99}{9}$

x< 11 $\displaystyle x< 11$

The correct solution set is $\left \{ x| x< 11 \right \}$ $\displaystyle \left \{ x| x< 11 \right \}$ .

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Example Question #81 : Algebra

Which of the following makes this equation true:

$\displaystyle 12x+4=112$

Possible Answers:

x=7 $\displaystyle x=7$

x=12 $\displaystyle x=12$

x=4 $\displaystyle x=4$

x=8 $\displaystyle x=8$

x=9 $\displaystyle x=9$

Correct answer:

x=9 $\displaystyle x=9$

Explanation:

To answer the question, we will solve for x. So, we get

$\displaystyle 12x+4=112$

$\displaystyle 12x+4-4=112-4$

12x=108 $\displaystyle 12x=108$

$\frac{12x}{12} = \frac{108}{12}$ $\displaystyle \frac{12x}{12} = \frac{108}{12}$

x=9 $\displaystyle x=9$

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Example Question #31 : Solving For The Variable

Solve for $\displaystyle x$ : $\displaystyle 3x+5 = -19$

Possible Answers:

$\displaystyle -17$

$\displaystyle 21$

$-\frac{14}{3}$ $\displaystyle -\frac{14}{3}$

$\displaystyle -6$

$\displaystyle -8$

Correct answer:

$\displaystyle -8$

Explanation:

Subtract five from both sides.

$\displaystyle 3x+5 -5= -19-5$

3x=-24 $\displaystyle 3x=-24$

Divide by three on both sides.

$\frac{3x}{3}=\frac{-24}{3}$ $\displaystyle \frac{3x}{3}=\frac{-24}{3}$

The answer is: $\displaystyle -8$

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Example Question #81 : Algebra

Solve the equation: $\displaystyle -2x-3 = 4x-7$

Possible Answers:

$-\frac{5}{3}$ $\displaystyle -\frac{5}{3}$

$\frac{3}{2}$ $\displaystyle \frac{3}{2}$

$\frac{5}{3}$ $\displaystyle \frac{5}{3}$

$\frac{2}{3}$ $\displaystyle \frac{2}{3}$

$\frac{5}{4}$ $\displaystyle \frac{5}{4}$

Correct answer:

$\frac{2}{3}$ $\displaystyle \frac{2}{3}$

Explanation:

Add $\displaystyle 2x$ on both sides.

$\displaystyle -2x-3 +2x= 4x-7+2x$

$\displaystyle -3 = 6x-7$

Add 7 on both sides.

$\displaystyle -3 +7= 6x-7+7$

4= 6x $\displaystyle 4= 6x$

Divide by 6 on both sides.

$\frac{4}{6}= \frac{6x}{6}$ $\displaystyle \frac{4}{6}= \frac{6x}{6}$

Reduce both fractions.

The answer is: $\frac{2}{3}$ $\displaystyle \frac{2}{3}$

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Example Question #41 : Solving For The Variable

Solve for x.

$\frac{x}{12} = 11$ $\displaystyle \frac{x}{12} = 11$

Possible Answers:

x=132 $\displaystyle x=132$

x=144 $\displaystyle x=144$

x=13 $\displaystyle x=13$

x=23 $\displaystyle x=23$

x=121 $\displaystyle x=121$

Correct answer:

x=132 $\displaystyle x=132$

Explanation:

To solve for x, we want x to stand alone. So, we get

$\frac{x}{12} = 11$ $\displaystyle \frac{x}{12} = 11$

$\frac{x}{12} \cdot 12 = 11 \cdot 12$ $\displaystyle \frac{x}{12} \cdot 12 = 11 \cdot 12$

x = 132 $\displaystyle x = 132$

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Example Question #42 : Solving For The Variable

Solve for $\displaystyle x$ : $\displaystyle 2x-5 = 3x-2$

Possible Answers:

$\displaystyle 3$

$-\frac{3}{5}$ $\displaystyle -\frac{3}{5}$

$\displaystyle -3$

$-\frac{7}{5}$ $\displaystyle -\frac{7}{5}$

$\frac{7}{5}$ $\displaystyle \frac{7}{5}$

Correct answer:

$\displaystyle -3$

Explanation:

Subtract $\displaystyle 2x$ from both sides.

$\displaystyle 2x-5 -2x= 3x-2-2x$

$\displaystyle -5 = x-2$

Add 2 on both sides.

$\displaystyle -5+2 = x-2+2$

-3 =x $\displaystyle -3 =x$

The answer is: $\displaystyle -3$

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Example Question #43 : Solving For The Variable

Which of the following makes this equation true:

$\displaystyle 7y+19=75$

Possible Answers:

y=19 $\displaystyle y=19$

y=9 $\displaystyle y=9$

y=7 $\displaystyle y=7$

y=12 $\displaystyle y=12$

y=8 $\displaystyle y=8$

Correct answer:

y=8 $\displaystyle y=8$

Explanation:

To answer the question, we will solve for y. We get

$\displaystyle 7y+19=75$

$\displaystyle 7y+19-19=75-19$

7y+0=56 $\displaystyle 7y+0=56$

7y=56 $\displaystyle 7y=56$

$\frac{7y}{7} = \frac{56}{7}$ $\displaystyle \frac{7y}{7} = \frac{56}{7}$

y=8 $\displaystyle y=8$

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Example Question #93 : Algebra

Solve the following equation: $\displaystyle 8x-15 = -3$

Possible Answers:

$\frac{3}{2}$ $\displaystyle \frac{3}{2}$

$\textup{The answer is not given.}$

$-\frac{3}{2}$ $\displaystyle -\frac{3}{2}$

$-\frac{9}{4}$ $\displaystyle -\frac{9}{4}$

$\frac{9}{4}$ $\displaystyle \frac{9}{4}$

Correct answer:

$\frac{3}{2}$ $\displaystyle \frac{3}{2}$

Explanation:

Add 15 on both sides.

$\displaystyle 8x-15+15 = -3+15$

Simplify both sides of the equation.

8x = 12 $\displaystyle 8x = 12$

Divide by 8 on both sides.

$\frac{8x}{8} = \frac{12}{8}$ $\displaystyle \frac{8x}{8} = \frac{12}{8}$

Reduce the fractions.

The answer is: $\frac{3}{2}$ $\displaystyle \frac{3}{2}$

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Example Question #94 : Algebra

Solve for $\displaystyle h$ : $\displaystyle 2h+6=4h+3$

Possible Answers:

$\frac{2}{3}$ $\displaystyle \frac{2}{3}$

$\frac{9}{2}$ $\displaystyle \frac{9}{2}$

$\frac{4}{3}$ $\displaystyle \frac{4}{3}$

$\frac{1}{3}$ $\displaystyle \frac{1}{3}$

$\frac{3}{2}$ $\displaystyle \frac{3}{2}$

Correct answer:

$\frac{3}{2}$ $\displaystyle \frac{3}{2}$

Explanation:

Subtract $\displaystyle 2h$ from both sides.

$\displaystyle 2h+6-2h=4h+3-2h$

6= 2h+3 $\displaystyle 6= 2h+3$

Subtract 3 from both sides.

$\displaystyle 6-3= 2h+3-3$

3=2h $\displaystyle 3=2h$

Divide by 2 on both sides.

$\frac{3}{2}=\frac{2h}{2}$ $\displaystyle \frac{3}{2}=\frac{2h}{2}$

The answer is: $\frac{3}{2}$ $\displaystyle \frac{3}{2}$

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All GED Math Resources

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GED Math : GED Math

Study concepts, example questions & explanations for GED Math

All GED Math Resources

Example Questions

Example Question #35 : Solving For The Variable

Example Question #32 : Solving For The Variable

Example Question #81 : Algebra

Example Question #31 : Solving For The Variable

Example Question #81 : Algebra

Example Question #41 : Solving For The Variable

Example Question #42 : Solving For The Variable

Example Question #43 : Solving For The Variable

Example Question #93 : Algebra

Example Question #94 : Algebra

All GED Math Resources

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